Question

1. Rezolvati ecuatia :
Ι Ιx-3Ι-5Ι=4
2. Rezolvati in Ζ ecuatia: (2x-3 )² = 25
Va rog sa ma ajutati !!!

Answer 1

1)
||x-3|-5|=4
|x-3|-5∈{4,-4}
|x-3|∈{9,1}
x-3∈{9,-9,1,-1}
x∈{12,-6,4,2}

2)(2x-3 )² = 25|√
|2x-3|=5
2x-3∈{5,-5}
2x∈{8,-2}
x∈{4,-1}

Answer 2

1.
Vom folosi faptul ca : 
ΙΕ(x)Ι= a (a≥0) inseamna Ε(x) = a sau Ε(X)=-a
Luand in considerare modulul exterior avem : Ιx-3Ι-5=-4
1) Ιx-3Ι-5=4; Ιx-3Ι=9 . De aici rezulta x-3=9 sau x-3=-9 si apoi x=12 sau x=-6 
2) Ιx-3Ι -5 = -4 ; Ιx-3Ι=1 . Deci : x-3=1; x=4 sau x-3 =-1; x=2 

2. Se obtine : Ι2x-3Ι = 5 ; deci 2x- 3 = 5 sau 2x-3 =-5 
2x-3=5 ; 2x=8; x=4 ;
 2x-3=-5 ; 2x=-2; x=-1
Succes!!!